solve: $z^3=\sqrt(3)-i$

Solution 1:

$$z^3=\sqrt{3}-i \Longleftrightarrow$$ $$z^3=\left|\sqrt{3}-i\right|e^{\arg\left(\sqrt{3}-i\right)i} \Longleftrightarrow$$ $$z^3=\sqrt{\left(\sqrt{3}\right)^2+1^2}e^{\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)i} \Longleftrightarrow$$ $$z^3=\sqrt{3+1}e^{-\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)i} \Longleftrightarrow$$ $$z^3=\sqrt{4}e^{-\frac{\pi}{6}i} \Longleftrightarrow$$ $$z^3=2e^{-\frac{\pi}{6}i} \Longleftrightarrow$$ $$z=\left(2e^{\left(2\pi k-\frac{\pi}{6}\right)i}\right)^{\frac{1}{3}} \Longleftrightarrow$$ $$z=2^{\frac{1}{3}}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} \Longleftrightarrow$$ $$z=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} $$

With $k\in\mathbb{Z}$ and $k:0-2$

So the solutions are:

$$z_0=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 0-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{\pi}{18}i}\approx 1.2407-0.2187i$$ $$z_1=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 1-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{\frac{11\pi}{18}i}\approx -0.4309+1.1839i$$ $$z_2=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 2-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{13\pi}{18}i}\approx -0.809-0.965i$$